Producing Exponential Generating Functions For Coefficients Exponential generating functions, becomes very complex and i just got the hang of non exponential function. And i am trying to apply the same methods that i used for regular generating functions. for example: the number of way to make n dollars among 5 people would be $(\frac{1}{1-x})^5$ . Now for exponential generating function for the number of words of length $n$ using letters $a, b, c, d, e$ such that all five letters occur with no restrictions 
Im wondering if anyone can assist with starting out the generating function for letter with word length $n$ would it be the generating function of $e^{5x}$
And then for 10 letters ill expand and look for the $\frac{x^{10}}{10!}$ coefficient
the following question after no restrictions is that at least three of the letters appear?
$(e^x-1)^3e^{2x}$ .. is  this generating function correct for at least three letters appear ?
 A: Your proposed answer, $(e^x-1)^3e^{2x}$, is the exponential generating function for the problem of counting the number of words composed of the letters a, b, c, d, e in which a, b, and c each appear at least once.  But I believe your problem is different: at least three different letters must appear, but they can be any three of the letters.
One way to approach the problem is to break it down into three cases:


*

*exactly three different letters appear,

*exactly four different letters appear,

*exactly five different letters appear.


The generating functions for the three cases are $10(e^x-1)^3$, $5(e^x-1)^4$, and $(e^x-1)^5$.  The coefficients are the numbers of ways of determining the set of letters that appear: $10=\binom{5}{3}$, $5=\binom{5}{4}$, $1=\binom{5}{5}$.
Summing these gives
$$
\begin{aligned}
(e^x-1)^3((e^x-1)^2+5(e^x-1)+10)&=(e^x-1)^3(e^{2x}+3e^x+6)\\
&=e^{5x}-10e^{2x}+15e^x-6.
\end{aligned}
$$
This last expression can be understood using the principle of inclusion-exclusion: $e^{5x}$ is the exponential generating function for words composed of the letters a, b, c, d, e.  We must exclude words that use two or fewer of the five letters.  We define $S_{a,b}$ to be the set of words composed of a and b, and similarly for other letter pairs.  The exponential generating function for $S_{a,b}$ is $e^{2x}$; that there are $10$ letter pairs accounts for the $-10e^{2x}$ term.  Words composed of only a single letter have been subtracted too many times, and must be added back.  Words composed only of the letter a, for example, have been subtracted four times, so we add them back three times, and similarly for the other four letters.  This accounts for the $+15e^x$ term.  Finally, the empty word has been added once, subtracted $10$ times, and added back $15$ times. Since the empty word does not use three letters, we need the $-6$ term to produce the correct count of $0$ for the empty word.
One can also achieve this result by doing a detailed analysis of the exponential generating functions for the intersections $S_{w,x}\cap S_{y,z}$, and similarly for three-, four-, etc. up to 10-set intersections.  Of the $\binom{10}{2}=45$ two-set intersections, for example, $30$ have exponential generating function $e^x$ ($S_{a,b}\cap S_{a,c}$, for example), while $15$ have exponential generating function $1$ ($S_{a,b}\cap S_{c,d}$, for example).  Similarly, $20$ of the three-set intersections have exponential generating function $e^x$ and $100$ have exponential generating function $1$, five of the four-set intersections have exponential generating function $e^x$ and $205$ have exponential generating function $1$, and all of the five- and higher set intersections have exponential generating function $1$.  After all additions and subtractions have been performed, we obtain the result above.
